Question

$f:A\to B$ is a function defined as $f\left(x\right)=|x^2+1|.\text{If}A=\{-2,,0,1,2,3,4\},$ then the possible set $B$ could be:

• A. $\{1,2,5,10,17,20\}$
• B. $\{1,2,5,10,15,16\}$
• C. $\{1,2,3,..,15\}$
• D. $\{1,2,5,10\}$

Answer 1

The correct option is A $\{1,2,5,10,17,20\}$

$f\left(x\right)=|x^2+1|$

So, $f(-2)=f\left(2\right)=5,f\left(0\right)=1,f\left(1\right)=2,f\left(3\right)=10,f\left(4\right)=17$
So, the range of the given function $=\{1,2,5,10,17\}$
$\because$ Set $B$ can be any set which contains all the elements of the range. i.e. Set $B$ can be any set which is the superset of the set (or equal to the set) containing all the elements of range.
$\therefore$ From the given options $B=\{1,2,5,10,17,20\}$